# Straight Lines (Circumcenter) Problems

### Straight Lines (Circumcenter) Problems

1) Find the circumcenter of the triangle whose vertices are (1, 0), (-1, 2) and (3, 2).

Solution: Vertices of the triangle are A (1, 0), B (-1, 2), C (3, 2). Let S (x, y) be the circumcenter of triangle ABC,

Then SA = SB = SC,

Let SA = SB ⇒ SA² = SB²,

(x – 1)² + y² = (x + 1)²+ (y – 2)²,

⇒ x² – 2y + 1 + y² = x² + 2x + 1 + y² – 4y + 4.

⇒ 4x – 4y = -4.

⇒ x – y = -1 … (1)

SB = SC ⇒ SB² = SC²

(x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²

⇒ x² + 2x + 1 + y² – 4y + 4 = x² – 6x + 9 + y² – 4y + 4

2x + 1 = -6x + 9

8x = 8 ⇒ x = 1

From equation (1)

x – y = – 1

1 – y = -1

y = 2,

Therefore circumcenter (x, y) = (1, 2).

2) Find the circumcenter of the triangle formed by the straight lines x + y = 0, 2x + y + 5 = 0 and x – y = 2.

Solution: Let the equation of AB be x + y = 0 … (1)

BC be 2x + y + 5 = 0 … (2)

And AC be x – y = 2 … (3) Solving (1) and (2), vertex B = (-5, 5)

Solving (2) and (3), vertex C = (-1, -3)

Solving (1) and (3), vertex A = (1, -1)

Let S (x, y) be the circumcenter of triangle ABC

Then SA = SB = SC

SA = SB

SA² = SB²

(x + 5)² + (y – 5)² = (x + 1)² + (y + 3)²

8x – 16y = -40

x – 2y = -5 … (4)

SA = SC

SA² = SC²

(x + 1)² + (y + 3)² = (x – 1)² + (y + 1)²

4x + 4y = -8

x + y = -2 … (6)

solving (4) and (5) point of intersection is (-3, 1)

circumcenter is S (-3, 1).